Integrala

Semaka · Mesaj de **Semaka** » 21 Noi 2020, 17:58

Buna seara!
$\int \frac{dx}{4-3sin^{2}x}$

Felixx · Mesaj de **Felixx** » 22 Noi 2020, 16:39

Facem substitutia de variabila:
$tgx=u$
$du=\frac{1}{cos^{2}x}dx=\left ( 1+tg^{2}x \right )dx=\left ( 1+u^{2} \right )dx\Rightarrow dx=\frac{du}{1+u^{2}}$
$sin^{2}x=\frac{tg^{2}x}{1+tg^{2}x}=\frac{u^{2}}{1+u^{2}}$
Atunci :
$I=\int \frac{1+u^{2}}{\left ( 1+u^{2} \right )\left ( 4u^{2}-3u^{2}+4 \right )}du=\int \frac{1}{u^{2}+2^{2}}du=...=\frac{1}{2}arctg\left ( \frac{tx}{2} \right )+c$

Semaka · Mesaj de **Semaka** » 22 Noi 2020, 17:25

Multumesc foarte mult

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